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Calculate the average kinetic energy of oxygen molecule at `0^(@)`C. `(R=8.314 J mol^(-1)` `K^(-1),N_(A)`=`6.02xx10^(23))`

Text Solution

Verified by Experts

Oxygen is diatomic molecule, therefore it has 5 degrees of freedom, 3 translational 2 rotational
`thereforeK.E.=(5)/92)K_(B)T`
`K.E.=(5)/(2)(RT)/(N_(A))`
`T=0^(@)C=273K`
`=(5)/(2)xx(8.314)/(6.023xx10^(23))xx273`
`=9.4xx10^(-21)J`
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